How do you prove that ${sin}^{2} (\frac{x}{2}) = \frac{1 - {cos}^{2} x}{2 (1 + cos x)}$ $sin^2(x/2) = (1-cos^2x)/(2(1+cosx))$ ?

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Answer 1 · 2016-01-09T14:48:16

This is not a valid identity.

Working only with the right hand side:

$\frac{1 - {cos}^{2} x}{2 (1 + cos x)}$ $(1-cos^2x)/(2(1+cosx))$

Use the Pythagorean Identity in the numerator.

$= \frac{{sin}^{2} x}{2 (1 + cos x)}$ $=sin^2x/(2(1+cosx))$

Split apart the fraction.

$= (\frac{sin x}{2}) (\frac{sin x}{1 + cos x})$ $=(sinx/2)(sinx/(1+cosx))$

Use the tangent half angle identity.

$= \frac{1}{2} sin x tan (\frac{x}{2})$ $=1/2sinxtan(x/2)$

There is nothing you could do here to make this equal ${sin}^{2} (\frac{x}{2})$ $sin^2(x/2)$ .

How do you prove that sin2(x2)=1−cos2x2(1+cosx) sin^2(x/2) = (1-cos^2x)/(2(1+cosx)) ?